What is inverse oflaplace 5/((s^2+5)) ?

The answer to inverse oflaplace 5/((s^2+5)) is sqrt(5)sin(sqrt(5)t)

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inverse oflaplace 5/((s^2+5))

inverse laplace

\frac{5}{( s ^{2} + 5 )}

5 sin (5 t)

Solution steps

L^{- 1} {\frac{5}{( s ^{2} + 5 )}}

= L^{- 1} {5 \frac{5}{s ^{2} + ( 5 ) ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 5 L^{- 1} {\frac{5}{s ^{2} + ( 5 ) ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

L^{- 1} {\frac{5}{s ^{2} + ( 5 ) ^{2}}} = sin (5 t)

= 5 sin (5 t)

\int \frac{4 ln ^{2} ( x ) + 49}{x} d x

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\frac{d}{d x} (\frac{7 e ^{x} + 4 e ^{- x}}{8})

\frac{d}{d x} (ln (\frac{( x - 4 )}{x - 3}))

\int - 2 e^{- t} d t

What is inverse oflaplace 5/((s^2+5)) ?
The answer to inverse oflaplace 5/((s^2+5)) is sqrt(5)sin(sqrt(5)t)